Seek help regarding the composition of coke oven flue gas concentrations

SO2≤50mg/Nm3, NO2≤80mg/Nm3

Based on the composition of the gas and the chemical reaction equations with oxygen, the theoretical amount of oxygen (and air) required, as well as the amounts and compositions of the products after combustion, are calculated as follows:

| Component | Reaction Equation | Theoretical Oxygen Consumption (m³/m³ gas) |
|-----------|-------------------|------------------------------------------|
| VCO₂ | | |
| VH₂O | H₂ + 0.5O₂ = H₂O | 0.53 |
| | | 1.46 |
| CH₄ | | 21.89 |
| | CH₄ + 2O₂ = CO₂ + 2H₂O | 243.78 |
| | | 43.78 |
| CO | | 5.51 |
| | CO + 0.5O₂ = CO₂ | 0.52 |
| | | 2.76 |
| 5.51 | CnHm | 1.86×0.8 |
| C₂H₄ | C₂H₄ + 3O₂ = 2CO₂ + 2H₂O | 34.46 |
| | | 2.98 |
| | | 2.98 |
| 1.86×0.2 | C₆H₆ | 7.52 |
| | C₆H₆ + 7.5O₂ = 6CO₂ + 3H₂O | 2.79 |
| | | 2.23 |
| | | 1.12 |
| CO₂ | | 2.13 |
| O₂ | | 0.49 |
| | | –0.49 |
| N₂ | | 5.22 |
| H₂O | | 2.35 |

Theoretical values for oxygen consumption and product amounts:
84.76, 34.74, 113.14, 5.22

Actual amounts of air, oxygen, and nitrogen introduced:
L_actual = α × L_theoretical = α × O_theoretical × (100/21)
1.3 × 84.76 × (100/21) = 524.70
524.70 × 0.0235 × 0.6 = 7.40
524.70 × 0.79 = 414.51
524.70 × 0.21 – 84.76 = 25.43

Amounts of various components in the exhaust gas, in m³:
34.74, 120.54, 419.73, 25.43, 600.44

Volume percentages of components in the exhaust gas:
5.79%, 82.08%, 69.90%, 4.24%, 100.00%

Note: CnHm is calculated assuming 80% C₂H₄ and 20% C₆H₆. The saturation temperature of the gas is 20°C, with a relative humidity of 0.6. The air excess factor α is 1.3

Regarding learning this*, I think it’s hard to determine; the composition of flue gas changes depending on the type of coal.

May I ask under what conditions (temperature, pressure, etc.) the actual flue gas volume and components were determined for Building 3#? Thank you!