I would like to ask everyone for help with a calculation problem related to sulfur recovery exhaust emissions
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I would like to ask everyone for help with a calculation related to sulfur recovery and exhaust gas emissions: For a sulfur processing plant with an annual production capacity of 100,000 tons, the overall sulfur recovery rate is designed to be 99.8%. To meet the requirements of the GB31579 environmental standard, flue gas alkaline washing technology is used to reduce the sulfur dioxide emission concentration to below 100 mg/m3, achieving an overall sulfur removal rate of 99.99%. What is the consumption rate of 30% sodium hydroxide per hour? How much wastewater containing 8% sodium sulfate is discharged? Thank you all.
This post was last edited by cg424 on 2022-10-25 at 11:29. The yield is 99.8% for 100,000 tons; the amount of sulfur emitted is about 200 tons per year, which corresponds to approximately 0.6 tons of sulfur per day. Using sodium hydroxide for alkaline washing and based on the production of sodium sulfide, two units of sodium hydroxide are required for each unit of sulfur, in a ratio of 32:80. Therefore, 0.6 tons of sulfur requires 1.5 tons of sodium hydroxide. When converted to a 30% solution, this amounts to 5 tons per day, or about 200 kilograms per hour. 2 units of sodium hydroxide produce 1 unit of sodium sulfite; the reaction ratio is 80:142. If all is converted to sodium sulfite, the amount would be around 2.7 tons per day. At a concentration of 8%, this amounts to approximately 33.8 tons per day, or 1.4 tons per hour.
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It is related to the controlled pH value; a high pH value causes part of the carbon dioxide to be absorbed and converted into sodium bicarbonate.
It should be GB31570-2015 \"Emission Standards for Pollutants from Petrochemical Enterprises\". As for the calculation, it’s actually quite simple – just perform a material balance. Give it a try.