Regarding the issue of coke oven flue gas volume
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I would like to ask fellow shipowners: There is a coke oven plant in a certain location, consisting of two sets of 5.5m rammed coke ovens, with an annual coke production capacity of 1 million tons. So, what is the volume of flue gas from the coke ovens (from the two chimneys), in cubic meters per hour? If waste heat recovery is used to produce steam at 0.7 MPa, how many tons can be generated per hour?
Do you need to design flue gas desulfurization?
Coke oven flue gas volume (flue gas volume from two chimneys; ca refers to the volume of gas returning to the furnace, plus an excess factor)
This can be calculated. First, tell me what your gas consumption is, what your A value is, and how long the coking time is; I’ll help you calculate it~~
Each unit of wet exhaust gas amounts to around 70,000; there is 4 tons of steam, and the temperature of the steam is closely related to that of the flue gases
The two coke oven chimneys produce approximately 10 tons of low-pressure steam at 0.7 MPa per hour.
Data such as the hourly gas consumption and the air excess factor are required for the calculation.
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 of products after combustion, are calculated. The composition of the gas is as follows: Content, Reaction Equation, Theoretical Oxygen Consumption. VCO2: 62.91; H2+0.5O2=H2O; 0.5; 31.46. CH4: 21.89; CH4+2O2=CO2+2H2O; 2; 43.78. CO: 5.51; CO+0.5O2=CO2; 0.5; 2.76. CnHm: 1.86×0.8; C2H4+3O2=2CO2+2H2O; 3; 4.46. Also, 1.86×0.2; C6H6+7.5O2=6CO2+3H2O; 7.5; 2.79. CO2: 2.13; 2.13. O2: 0.49; -0.49. N2: 5.22; 5.22. H2O: 2.35. Theoretical values for oxygen consumption and product amounts are 84.76, 34.74, 113.14, and 5.22 respectively. The actual amounts of air, oxygen, and nitrogen used are calculated as follows: L_actual = α×L_theoretical = α×O_theoretical×(100/21). Thus, 1.3×84.76×(100/21) = 524.70. Then, 524.70×0.0235×0.6 = 7.40. 524.70×0.79 = 414.51. And 524.70×0.21 – 84.76 = 25.43. The amounts of various components in the exhaust gas, in m3, are 34.74, 120.54, 419.73, 25.43, totaling 600.44 m3. The volume percentages of these components in the exhaust gas are 5.798%, 20.08%, 69.90%, 4.24%, adding up to 100.00%. Note: CnHm is calculated based on 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