【Professional】Encyclopedia of wastewater treatment equipment, 300 questions on wastewater treatment! ! (Uniquely complete)

【Excessive ammonia nitrogen】 Question: Our factory uses an anaerobic-hydrolysis-first-stage aerobic contact oxidation-second-stage aerobic contact oxidation process. The influent COD is below 1000 mg/L ； Ammonia nitrogen in the inlet water: 50 mg/L ； The BOD5/COD ratio is above 0.35. The ammonia nitrogen level in the effluent does not meet the standards; how to solve this? Expert’s answer: Your process needs to be changed; this approach will not allow you to meet the required standards. With an ammonia nitrogen level of 50 mg/L in the incoming water (and total nitrogen levels even higher), there is no need for hydrolysis or acidification when the BOD5/COD ratio is above 0.35. Nor is anaerobic treatment necessary when the COD level is below 1000 mg/L. Both the anaerobic tank and the hydrolysis tank can be replaced with aerobic tanks. There is no need to set up a separate denitrification tank either; it is sufficient to keep the dissolved oxygen level in the current first-stage aerobic contact oxidation tank below 0.5 (this assumes that both the hydrolysis tank and the anaerobic tank are replaced with aerobic tanks). Since I am not aware of the specific details in this regard, this is only a preliminary idea.

Are there any good methods for dealing with alkaline slag?

【Excessive ammonia nitrogen】Following up on the previous question: Why is it said that hydrolysis and acidification are not necessary when BOD5/COD is above 0.35? Expert’s answer: Since wastewater with such a B/C ratio has decent biodegradability, the amount of non-biodegradable substances in it is not high at this ratio; most of these substances can be adsorbed by activated sludge and removed through the excess sludge, thereby ensuring that the effluent meets the required standards. It should also be noted that some of the so-called non-biodegradable organic substances can still be degraded, only the biochemical process takes a longer time. When I say there’s no need for acidification, it’s not because the acidification effect is poor, but rather from economic considerations such as investment and land use.

【Urban Sewage Treatment – CAST Process】 Question: When using the CAST process to treat urban sewage, with a BOD level of around 80 mg/L and an MLSS level of around 4000 mg/L, the DO level is currently maintained between 1.0 and 3.0 mg/L during the treatment process; however, it sometimes exceeds 3.0 mg/L. The ash content in the sludge is currently high; what specific aspects should be taken into consideration during restoration, and what are the approximate control parameters? What is wrong with the parameters above? Expert answer: Based on the information provided, it is likely that the low sludge load is causing the sludge to age; therefore, the amount of sludge discharged should be increased, the flow rate back to the selection tank should be reduced, and the aeration time should be shortened.

【Excessive COD】 Question: (1) Recent trials in the workshop have caused abnormal water inflow. Yesterday, the COD level was 6,000 mg/L, while the design value is only 600 mg/L. What measures should be taken to restore normal conditions of the effluent as soon as possible? (2) Recently, the air pressure in the air compressor room was 8 kilograms, and since there was no pressure relief valve installed, they explained that the flow valve of the aeration pipe could also be used to control the pressure. May I ask if uneven aeration is caused by excessive wind pressure? Expert answer: If the COD of the incoming water is more than ten times the designed value, the standards cannot be met; it is necessary to increase the oxygen supply and reduce or eliminate sludge discharge, with the aim of controlling both the sludge load and the oxygen supply. But note: reducing or stopping sludge discharge is only temporary; after a period of inactivity (at least half a day), the sludge discharge rate should be increased again. 　　　The purpose of the above measures is to first mix the sludge with high-concentration wastewater for adsorption; after some time, part of the organic matter decomposes, but most of it remains adsorbed on the sludge, allowing it to be removed from the system along with the sludge. This helps the system return to normal more quickly, as high-concentration wastewater generally does not persist for very long. A wind pressure of 8 kilograms is not acceptable.

More answers: The pressure of fans used in wastewater treatment generally does not exceed 1 kg/cm2; the pressure increase required by the fan is determined based on the aeration depth of the tank. For a standard tank with a depth of 4 meters, a pressure increase of 0.05 MPa is sufficient, which helps to save energy.

I would like to ask experts: when using the parallel multi-effect evaporation process to treat sodium chloride wastewater, after the first stage of evaporation, two streams of feed liquid emerge from the evaporator – one stream goes to the second stage of evaporation, while the other is recycled back to the first stage. Generally speaking, what should be the ratio between these two streams?

One stream enters the second-effect evaporator, while another is recycled back to join the raw water in the first-effect evaporator? It seems pretty complicated; it has to be calculated using thermodynamic equations, right? The basic condition for determining how to divide the proportions is that the system can operate properly. I hope someone can give some guidance

During simulation, you can choose whatever method you like; it will still converge