When cooling flue gas from 130°C to 60°C, how should the amount of water used for spraying be determined?
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Can the flue gas at 130°C be cooled to 60°C, assuming that the amount of water sprayed will cause complete vaporization? Seeking advice, thank you!
{:3_54:} That can’t be right. If you calculate it based on vaporization, the temperature of your cooling medium is already 100 degrees Celsius; so how can the temperature of your process medium drop to 60 degrees? If it doesn’t work, calculate it separately – first reduce the temperature of the flue gas to 100 degrees, and then reduce it from 100 to 60 degrees. Handle the moisture content as a separate factor
The actual water consumption is definitely a bit higher than this, but not by much
In adiabatic calculations, the heat released upon cooling the flue gas from 130 degrees Celsius is absorbed by the moisture, causing it to evaporate. The heat Q required for cooling the flue gas is equal to the change in enthalpy; it is necessary to calculate the enthalpy values at 130 degrees and 60 degrees based on the composition of the flue gas. The evaporation of water consists of two components: the sensible heat due to temperature rise, Cm△T, and the latent heat of vaporization, mr. Reference can be made to the relevant content on chemical engineering principles. This should be a relatively simple calorimetric calculation
The actual usage can be referenced from the cooling of coke oven raw gas, with the process being controlled by the rates of heat and mass transfer. Calculations can only be performed for heat balance.
It also depends on the temperature of the water source and the amount of smoke, right? If the flue gas contains water vapor, it will also liquefy, resulting in a higher amount of water needed for spraying.