How to calculate the amount of water needed to dilute COD?
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There is currently 36 cubic meters of wastewater with a COD concentration of 3000 mg/l. To reduce this COD concentration to 500 mg/l, how much clean water is needed for dilution (assuming the COD level in clean water is 0)? Is there any formula related to COD dilution?
A layperson might reply awkwardly: Can it be simply understood as the solute remaining unchanged? 36×3000÷500－36＝180 cubic
It’s very simple; the one upstairs is the correct one
In a hurry? Why not use the Fenton process, or add a small amount of acid? Just diluting with water isn’t sufficient to meet the standards