Flue gas volume of 6m coke oven?
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Our factory’s coke ovens are equipped with online monitoring systems. The volume of flue gas on the north side is 250,000 cubic meters (as stated); what should it be?
Are you the smoke output of a furnace? What are the operating conditions?
By providing the furnace type (size of the carbonization chamber, number of holes) and the cycle time, it is possible to calculate the gas consumption per hour, that is, the gas flow rate. By combining this with the air excess factor, it is possible to determine the exhaust gas flow rate, that is, the amount of exhaust gas produced per hour
Do you know how to calculate the amount of exhaust gas based on the gas consumption?
Detailed calculation method: 1. First, determine the percentage (by volume) of the components in your coke oven gas, such as: H2, CH4, CO, CnHm, CO2, O2, N2. Their calorific values are 62.9, 12.19, 5.51, 1.86, 2.13, 0.49, 5.22; the total calorific value is 16690.25. 2. Based on the composition of the gas and the chemical reaction equations with oxygen, calculate the theoretical amount of oxygen (and air) required, as well as the amounts of waste gases produced after combustion.
| Component | Content (%) | Reaction Equation | Theoretical Oxygen Consumption (m3/m3 of gas) |
|-----------|--------------|-------------------|--------------------------------------------|
| VCO2 | 62.9 | H2 + 0.5O2 = H2O | 0.5 × 31.46 = 15.73 |
| VH2O | 12.19 | CH4 + 2O2 = CO2 + 2H2O | 2 × 33.78 = 67.56 |
| VO2 | 5.51 | CO + 0.5O2 = CO2 | 0.5 × 2.76 = 1.38 |
| V CnHm | 1.86 | C2H4 + 3O2 = 2CO2 + 2H2O | 3 × 34.46 = 103.38 |
| | 1.86×0.2 | C6H6 + 7.5O2 = 6CO2 + 3H2O | 7.5 × 2.23 = 16.73 |
| CO2 | 2.13 | — | 2.13 |
| O2 | 0.49 | — | 0.49 × 0.49 = 0.2401 |
| N2 | 5.22 | — | 5.22 |
| H2O | 2.35 | — | 2.35 |
Theoretical values for oxygen consumption and waste gases: 84.76 m3, 34.74 m3, 113.14 m3, 5.22 m3 respectively.
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 m3
524.70 × 0.0235 × 0.6 = 7.40 m3
524.70 × 0.79 = 414.51 m3
524.70 × 0.21 – 84.76 = 25.43 m3
Amounts of various components in the waste gases, in m3:
34.74, 120.54, 419.73, 25.43, 600.44 m3
Volume percentages of components in the waste gases:
5.79%, 82.08%, 69.90%, 4.24%, 100.00%
Note: CnHm is calculated as 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 1.33. Assuming that each coke oven consumes 9900 m3/h of gas, the air flow entering the coke oven is 9085 × 5.247 = 47669 m3/h, while the waste gas flow is 9085 × 6.0044 = 54550 m3/h.
| Component | Content (%) | Reaction Equation | Theoretical Oxygen Consumption (m3/m3 of gas) |
|-----------|--------------|-------------------|--------------------------------------------|
| VCO2 | 62.9 | H2 + 0.5O2 = H2O | 0.5 × 31.46 = 15.73 |
| VH2O | 12.19 | CH4 + 2O2 = CO2 + 2H2O | 2 × 33.78 = 67.56 |
| VO2 | 5.51 | CO + 0.5O2 = CO2 | 0.5 × 2.76 = 1.38 |
| V CnHm | 1.86 | C2H4 + 3O2 = 2CO2 + 2H2O | 3 × 34.46 = 103.38 |
| | 1.86×0.2 | C6H6 + 7.5O2 = 6CO2 + 3H2O | 7.5 × 2.23 = 16.73 |
| CO2 | 2.13 | — | 2.13 |
| O2 | 0.49 | — | 0.49 × 0.49 = 0.2401 |
| N2 | 5.22 | — | 5.22 |
| H2O | 2.35 | — | 2.35 |
Theoretical values for oxygen consumption and waste gases: 84.76 m3, 34.74 m3, 113.14 m3, 5.22 m3 respectively.
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 m3
524.70 × 0.0235 × 0.6 = 7.40 m3
524.70 × 0.79 = 414.51 m3
524.70 × 0.21 – 84.76 = 25.43 m3
Amounts of various components in the waste gases, in m3:
34.74, 120.54, 419.73, 25.43, 600.44 m3
Volume percentages of components in the waste gases:
5.79%, 82.08%, 69.90%, 4.24%, 100.00%
Note: CnHm is calculated as 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 1.33. Assuming that each coke oven consumes 9900 m3/h of gas, the air flow entering the coke oven is 9085 × 5.247 = 47669 m3/h, while the waste gas flow is 9085 × 6.0044 = 54550 m3/h.