4. During operation, a thick layer of sludge accumulates on the surface of the oxidation ditch, with sludge particles of about 1 mm in diameter appearing yellowish. This often causes a large amount of sludge floating in the secondary sedimentation tank, sludge whitening, flocs flowing out with the effluent, a rapid decrease in SV30, and loss of treatment effect. The accumulated sludge is thinned and eliminated, and the cycle repeats. Please explain the cause and control measures.

Answer: This indicates that the sludge has lost its activity, increasing ESS. There are two possibilities: one is sludge self-oxidation; the other is sludge poisoning. Judging from the phenomena you described, the former is more likely. The specific oxygen consumption rate, that is, the ratio of endogenous oxygen consumption rate to substrate oxygen consumption rate, can be measured to determine the cause and take targeted measures.

5. How is section A of the AB process controlled? Is it continuously recycled to section A from the primary sedimentation tank at an equal flow rate? What should the SV30 be controlled at? Is it 5%-10%?

Answer: The recycle ratio of section A should be higher, but the sludge retention time in the primary sedimentation tank should not be too short. Although section A mainly focuses on adsorption, it also has a certain degree of biodegradation. Most biodegradation occurs in the sedimentation tank. Only by degrading the organic matter adsorbed on the sludge surface can the adsorption capacity be restored. MLSS should be used for control. When the sludge settling performance is stable, SV30 can also be used. It should be determined based on the actual situation. A settling ratio of 5%-10% is too low.

6. If a wastewater treatment plant has not achieved optimal treatment results after one or two years of operation, should it consider re-culturing (sludge replacement)? What is the difference between sludge replacement and initial culturing?

Answer: No need to replace! If the operating conditions remain unchanged, replacement will yield the same results. Even if you add dominant strains, it will only last for a while. The key is to control the operating conditions well. If there are design problems, they should be rectified promptly.

7. I am commissioning industrial wastewater. The process is hydrolysis + anaerobic + aerobic tank 1 + aerobic tank 2 + sedimentation. Due to installation problems, the aeration in the aeration tank is uneven (circular aeration head aeration). At each aerator, there is a fountain-like up-and-down rolling (diameter of about 1 m), uneven aeration. How much does this affect the treatment effect? It was also found that there is less biofilm on the filler in the aeration zone, and microscopic examination revealed large metazoans, but no other organisms were found. The surface of the biofilm on the filler is light yellow, and the biofilm outside the aeration zone is 3 cm thick. Can you explain this?

Answer: The situation you described cannot be considered uneven aeration; it is a normal phenomenon. Also, you mentioned that the biofilm is not abundant, but I don't know how much. If the biofilm basically covers the filler, it's good. As for the biofilm outside the aeration zone being 3 cm thick, it means serious balling, and measures should be taken, such as high-volume flushing and anaerobic biofilm removal.

8. Please answer the following questions about the contact oxidation tank.

(1) How long can the sludge on the filler survive when the contact oxidation tank is emptied?

(2) Should nutrients be added when the treatment capacity of the contact oxidation tank decreases?

(3) Is adding kerosene to control foam effective? If so, how much is usually added?

Answer: The answers to the three questions are as follows:

(1) When the contact oxidation tank is emptied, it's not a matter of how long the biofilm sludge can survive, but rather to prevent the soft filler from drying out and becoming caked. Once caked, it is difficult to spread out again when immersed in water. This situation should be prevented.

(2) The decrease in the treatment capacity of the contact oxidation tank should be considered from multiple factors. The control of biofilm thickness is very important. A thick biofilm will seriously affect the treatment capacity. Also, note that the tank should be emptied slowly, otherwise, the soft filler racks with a large amount of biofilm will collapse or deform.

(3) Chemical foam is more effectively treated with water spray (do not directly flush with water). I do not agree with using methods such as kerosene to eliminate foam.

9. The average data for influent, effluent, and various biochemical pools in our plant over the past week is as follows:

Influent: BOD: 253; COD: 810; pH: 7.9; SS: 286; Color: 32 times; Ammonia nitrogen: 28; Total nitrogen: 64; Total phosphorus: 6.0;

Effluent: BOD: 4.8; COD: 74; pH: 8.1; SS: 12; Color: 8 times; Ammonia nitrogen: 7.6; Total nitrogen: 22.8; Total phosphorus: 1.02;

Biochemical pool: MLSS: 4200; MLVSS: 2340; SV%: 47.2; Sludge index: 118.9; Sludge age is 35 days.

The improved activated sludge process is used. The current influent is only about 25,000 tons/day (the design is 50,000 tons), more than 80% of which is industrial wastewater, and there is a small amount of high-concentration leachate. The process flow is aeration sedimentation tank - post-biochemical pool - secondary sedimentation tank, without contact tank and hydrolysis tank. The biochemical pool is air-supplied by a blower, deep-water rotary disc aeration. When the water continuously enters, the dissolved oxygen cannot reach 1mg/L, and after the water stops entering, the dissolved oxygen slowly rises to about 4-5mg/L. The serious excess of influent and the defects of the structure have resulted in a high load on the biochemical pool, and the sludge thickening pool is very small (180 cubic meters), with a considerable part of the excess sludge returning to the influent pump room.

The problems encountered now are: (1) After the influent enters the secondary sedimentation tank, suspended activated sludge particles are often found. Is this due to insufficient settling time or difficulty in sedimentation?

(2) Aggregations of red worms (water fleas) were found in all three secondary sedimentation tanks. Water fleas seem to be an indication of good water quality. Is it due to high sludge concentration leading to massive reproduction?

(3) A thin layer of floating sludge is sometimes found in the secondary sedimentation tank. Is this due to poor sludge settling performance, insufficient aeration in the biochemical pool, or untimely sludge return?

(4) Algae or moss easily grows on the weir plate of the secondary sedimentation tank. What methods can be used to overcome this?

(5) I believe that the sludge is severely aged. The MLSS should be controlled to between 3000-3500 or lower, the amount of excess sludge discharged should be increased, and the sludge age should be reduced. Will this reduce the shock resistance of the biochemical pool? Will the effluent water quality improve?

Answer: The sludge is somewhat aged, but not very serious. The sludge age has reached 35 days. Based on this calculation, the sludge load is less than 0.03. Controlling the current sludge concentration to 2/3 is sufficient. The sludge concentration should be gradually reduced. Water fleas have no effect on the effluent. Do not collect water fleas when analyzing samples. Attention should also be paid to the control of the sedimentation tank sludge layer. Algae and moss on the weir plate of the secondary sedimentation tank can only be manually removed.

10. We use two-stage biochemical treatment for petrochemical wastewater. The first stage is a circular completely mixed aeration tank, and the second stage is a push-flow aeration tank. The DO in the first stage is 0.2mg/L, and the DO in the second stage is 5.0mg/L. During this period, the pH of the influent in the first stage of biochemical treatment is 8.0, the effluent is 6.5, and the pH after the second stage of biochemical treatment is 5.78, which exceeds the range of 6-9. What is the reason?

Answer: Low DO in the first stage is normal because the sludge load is high. The reason for the decrease in pH in the first stage may be acidification due to excessive load. The decrease in pH of the effluent in the second stage may be caused by the consumption of alkalinity due to nitrification. Because your introduction is too simple, I can only make a simple analysis and inference.

11. For ammonia nitrogen removal, in addition to sufficient carbon sources, a sufficiently long sludge age, and sufficient return flow, is the return flow the return flow of the effluent from the aerobic pool or the bottom return flow of the secondary sedimentation tank? I am currently debugging nylon wastewater. The original design was to return the effluent from the aerobic pool, but in fact, if the return flow rate reaches twice the amount, the anaerobic environment in the preceding anoxic pool cannot be guaranteed. My master said that it would be better to control the dissolved oxygen in the aerobic pool at around 1mg/L. Is this correct?

Answer: Based on your introduction, it should be pre-denitrification, which requires the return flow of the effluent from the aerobic pool and the sludge from the secondary sedimentation tank. It is inappropriate to say that if the return flow rate reaches twice the amount, the anaerobic environment in the preceding anoxic pool cannot be guaranteed. The anoxic zone is not equal to anaerobic. DO less than 0.5mg/L is sufficient. Your master's statement that the dissolved oxygen in the aerobic pool should be controlled at around 1mg/L is also reasonable, which can prevent the DO in the anoxic zone from being greater than 0.5mg/L.

If the DO in the aerobic zone is around 1, and the effluent return flow rate is twice the amount, and the DO in the anoxic zone is still greater than 0.5mg/L, the dissolved oxygen in the aerobic zone should not be further reduced, and the effluent return flow rate (the amount of nitrate nitrogen entering the anoxic zone will be less) should not be arbitrarily reduced. At this time, without affecting the effect of mud-water separation in the secondary sedimentation tank, the amount of sludge discharged from the secondary sedimentation tank can be reduced to increase the sludge layer in the tank, so that the residence time of the sludge in the secondary sedimentation tank is increased, so that it is in a state of oxygen deficiency or anoxia, which is also conducive to preventing the DO in the anoxic zone from rising.

Reducing the amount of sludge discharged from the secondary sedimentation tank will not affect the amount of sludge returned to the reaction tank, because when the sludge layer in the secondary sedimentation tank is increased, the concentration time of the sludge in the sludge layer is longer. In this case, the amount of sludge discharged is reduced, but the concentration of the discharged sludge is increased. If it is a contact oxidation process, the effluent needs to be returned, and the sludge will not be returned. I do not agree with using pre-denitrification. Because the energy consumption of effluent return flow is large, a large return flow rate requires a large reaction tank volume. The statement about the removal of nitrifying bacteria is inappropriate, but I understand your meaning.

12. (1) Recent plant commissioning has resulted in abnormal influent. Yesterday, the COD was 6000, while the design was only 600. What measures should be taken to restore the effluent to normal as soon as possible?

(2) Recently, the air pressure in the air compressor room is 8 kg, and no pressure reducing valve is installed. They explained that the flow valve of the aeration pipe can control the pressure. Please, is the uneven aeration caused by excessive air pressure?

Answer: If the influent COD is more than ten times the design value, it is impossible to meet the standard. The oxygen supply should be increased, and the amount of sludge discharged should be reduced or no sludge should be discharged. The purpose is to control the sludge load and oxygen supply. However, it should be noted that reducing or not discharging sludge is temporary. After a reaction interruption (at least half a day), the amount of sludge discharged should be increased.

The purpose of the above measures is to first mix and adsorb the sludge with high-concentration wastewater. After a period of time, some organic matter will degrade, but a large part of the organic matter will still be adsorbed onto the sludge and discharged from the system with the sludge. This allows the system to recover quickly, as high-concentration wastewater generally does not last for a long time. A wind pressure of 8 kg is definitely not feasible.

13. In the activated sludge process for treating fish processing wastewater, the biochemical process is carried out in three series of tanks. Currently, a large amount of foam has appeared in the second and third biochemical tanks, but there is no foam in the first biochemical tank. Initially, it was thought to be detergent foam, but recently, during the peak of detergent use, the water has been discharged for four or five days, and there is still no improvement, and there are signs of increase. What is the reason, and how to solve it?

Answer: It may be biological foam caused by Nocardia, which is easy to breed in the later stage with low oil and low load. This kind of foam is difficult to eliminate with water spraying, and can only be manually removed or allow some raw water to directly bypass to the later biochemical tank, which can suppress the reproduction of Nocardia to a certain extent.

14. The old device is modified to treat ammonia nitrogen wastewater. Hydrolysis + anaerobic + two-stage aerobic (contact oxidation process) is adopted. Wastewater is returned to the hydrolysis tank, and sludge is returned to the anaerobic tank (anoxic tank). If the reflux is increased, the sludge loss in the hydrolysis tank is very fast (the hydrolysis tank changes from black to clear), and the dissolved oxygen in the subsequent anaerobic tank can reach 0.7.

Therefore, it is attempted to return the flow from the bottom of the sedimentation tank (through the vent pipe). Due to the limitation of the return flow, the removal rate of ammonia nitrogen is not ideal. Question: In the pre-denitrification process, is the effluent from the aerobic tank or the sedimentation tank usually returned?

Answer: The effluent from the secondary aerobic tank should be returned to the anoxic zone, not to the hydrolysis tank and anaerobic tank. It may be that you haven't fully explained it clearly. I always feel that there is a problem with this process. The hydrolysis tank is the acidification tank, which mainly improves the biodegradability of wastewater through hydrolysis and acidification. The nitrification effect should be understood first, and then the denitrification problem should be considered. Also, is the sedimentation tank you mentioned the final sedimentation tank (used to settle the biofilm shed from the aerobic tank)? Is there a sedimentation tank after the anaerobic tank? I feel that in addition to design problems, there are also operational management problems.

15. Currently, SBR technology is used to treat hospital wastewater. Domestic sewage and return sludge (1000 catties of sludge from the belt sludge machine) have been added. When aeration is carried out, a large amount of white foam appears in about ten minutes. The water volume is about 120 cubic meters. Is the inflow large and the concentration high? What preparations are needed for the next step? How to cultivate microorganisms better? How to control the aeration time? How to solve this problem?

Answer: If dehydrated sludge is used for sludge cultivation and inoculation, the amount added should be at least 3% of the effective tank volume, and there are also nutritional requirements. The amount of inoculated sludge added is too small. As for the appearance of foam, it is normal, and it will be greatly reduced or disappear after sludge formation. The latter problems are specific operational control problems, which will not be elaborated here.

16. Our factory adopts anaerobic-hydrolysis-primary aerobic contact oxidation-secondary aerobic contact oxidation process. Influent COD is below 1000mg/L; influent ammonia nitrogen is 50mg/L; BOD 5 /COD is above 0.35. The effluent ammonia nitrogen cannot meet the standard. How to solve it?

Answer: Your process should be changed, it is impossible to meet the standard in this way. The influent ammonia nitrogen is 50mg/L (total nitrogen is even higher), and BOD5/COD is above 0.35, so hydrolysis and acidification are not necessary. Anaerobic is not necessary for COD below 1000mg/L. The anaerobic tank and hydrolysis tank can be changed to aerobic tanks (contact oxidation). The denitrification tank does not need to be set separately. As long as the dissolved oxygen in the current primary aerobic contact oxidation tank is controlled below 0.5 (assuming that the hydrolysis tank and anaerobic tank are changed to aerobic tanks), because the specific situation in various aspects is not yet understood, it is only a preliminary idea.

17. Why do you say "BOD 5 /COD above 0.35 does not require hydrolysis and acidification"?

Answer: Because the biodegradability of wastewater with such a B/C ratio is still good, the non-biodegradable substances in the wastewater are not very high at this ratio, and most of them can be adsorbed by activated sludge and removed through excess sludge discharge to meet the effluent standards. It should also be explained that some of the so-called non-biodegradable organic matter can still be degraded, but the biodegradation process takes a longer time. I said that acidification is not necessary, not because the acidification effect is not good, but from the perspective of investment, land occupation, and other economic aspects.

18. CAST process for treating urban wastewater, BOD is around 80, MLSS is around 4000mg/L, and the current DO is controlled at 1.0~3.0 during the reaction, and sometimes DO exceeds 3.0. Currently, the sludge ash content is high. What aspects should be paid attention to during recovery, and what are the approximate control parameters? What is wrong with the above parameters?

Answer: Based on the information provided, it may be that the sludge load is too low, causing sludge aging. The sludge discharge rate should be increased, the return flow rate to the selection tank should be reduced, and the aeration time should be reduced.

19. If the sulfide in wastewater is high and wet oxidation is used, what if sulfuric acid is generated? This will corrode the pipe wall, possibly causing the pipe wall to collapse. Is it better to let the sulfide precipitate?

Answer: There is no such problem as you said. In the wet oxidation method, sulfide is oxidized to sulfate, and of course, there will be some partially unoxidized thiosulfate.

20. What is the amount of dry sludge added directly related to? How much should be added for the initial cultivation?

Answer: How much sludge is needed for inoculation cultivation can only be a rough range, the key is still experience, otherwise, even if the most sludge is inoculated, it will be useless.

I once saw a post here, where a unit directly transplanted and cultivated and domesticated activated sludge from the chemical wastewater treatment equipment of a nearby similar factory. The amount of sludge transferred was very large, and a lot of sludge transportation costs were spent, but the cultivation and domestication failed after nearly a month. This was caused by improper control of the cultivation and domestication process.

21. We use A ² O process. Now the total phosphorus removal is acceptable, but the ammonia nitrogen has not been reduced. The debugging has been going on for three months. I once saw an article saying that ammonia nitrogen can be reduced without internal reflux, and our internal reflux is difficult to control, and there is almost none. I don't know what to do to reduce ammonia nitrogen?

Answer: Based on your description, the higher effluent ammonia nitrogen compared to influent is unrelated to the lack of recirculation. The main reason is insufficient reaction time. It is estimated that this type of wastewater has high organic nitrogen. Due to insufficient nitrification time, the ammonification rate of organic nitrogen is greater than the nitrification rate of ammonia nitrogen, resulting in a normal increase in effluent ammonia nitrogen. It is also necessary to confirm whether the basic conditions for nitrification are well controlled.

22. During the cultivation of biofilm in the contact oxidation device, it is found that the biofilm forms and then falls off. How to solve and avoid this?

Answer: It is a normal phenomenon that the biofilm forms and then mostly falls off. Generally, after the second or third time of re-formation, the biofilm is considered successfully attached. In other words, the first biofilm formation cannot be considered successful. If there is no large-scale shedding after the first biofilm formation, it is accidental. After one or two shedding, the formation is inevitable. In most cases, this is the case.

23. Acrylonitrile wastewater is difficult to treat. What treatment process is suitable?

Answer: Acrylonitrile wastewater has poor biodegradability and contains a large amount of oligomers and inorganic COD such as SCN. Therefore, pretreatment is required, such as neutralization and coagulation, followed by biological treatment. For biological treatment, biofilm method is recommended, with an acidification process beforehand.

24. Should the hydraulic retention time of the contact oxidation pool be calculated according to the void fraction of the filler? How to calculate it?

Answer: It is meaningless to calculate the hydraulic retention time according to the void fraction of the filler, and it is not accurate. It should be the volumetric load and the retention time of wastewater in the bioreactor.

25. Will COD increase in the hydrolysis acidification stage? I mean, macromolecules are hydrolyzed into small molecules. Originally, some macromolecules in the water could not be oxidized by potassium dichromate, but they can be oxidized after hydrolysis. I am treating leachate from garbage.

Answer: It is indeed possible that macromolecular organic matter that could not be oxidized by potassium dichromate before can be oxidized after hydrolysis and acidification, but the COD of the effluent from the hydrolysis acidification pool will not increase. The reason is:

(1) When determining COD by the potassium dichromate method, silver sulfate is used as a catalyst, which can oxidize more than 95% of organic matter;

(2) Part of the COD will also be removed during the hydrolysis and acidification process, and the removal rate is definitely higher than the substances that cannot be oxidized by potassium dichromate mentioned above.

26. (1) When we measure ammonia nitrogen by distillation titration, the distillate is yellow, which affects the titration endpoint. I don't know why, how to avoid or eliminate the interference.

(2) When measuring the concentration of aerobic sludge, is it 10 ml of sludge that has been settled for half an hour, or 10 ml of a mixture of water and sludge that has been settled? What is the normal range for the concentration of aerobic sludge?

(3) What is the normal sludge concentration in a hydrolysis acidification pool?

Answer: High concentration should be diluted and then determined by colorimetry. If the yellow color remains after adding the chromogenic agent, it indicates that the ammonia nitrogen concentration is very low (just a guess). The sludge concentration should be determined using 100 ml of mixed solution and the settled sludge in a graduated cylinder. The control range of sludge concentration should be determined according to the actual sludge load of the device, and cannot be generalized.

27. Question: How does the Carousel 2000 operate during the Spring Festival (some people go home for the Spring Festival and there is no shift work)?

Answer: As long as the wastewater continues, people cannot rest. The so-called weekend operation mode is unreliable.

28. Our plant's UNITANK system mainly consists of a three-compartment structure (the three compartments can be divided into the left compartment, the middle compartment, and the right compartment). The three compartments are interconnected, each compartment has an aeration system, using mechanical surface aeration, and equipped with stirring. The outer two compartments have weirs and sludge discharge devices. The two compartments alternately serve as aeration and sedimentation tanks, and wastewater can enter any of the three compartments. The current process operation is divided into two main operation stages:

(1) Wastewater first enters the left compartment, and the left compartment is anaerobically stirred for 1 hour. The middle compartment is aerated aerobically, and the right compartment is used as the effluent sedimentation tank.

(2) Wastewater continues to enter the left compartment. The left compartment stops stirring and is aerated aerobically for 3.5 hours. The middle compartment is always aerated aerobically, and the right compartment is still used as the effluent sedimentation tank.

(3) The left compartment stops aeration and settles for 1 hour. Wastewater is transferred from the left compartment to the middle compartment. The middle compartment is always aerated aerobically, and the right compartment is still used for effluent. After the first main operation stage (6 hours) is completed, after a short transition period (0.5 hours backwashing), the second main operation stage is entered. In the second main operation stage, the wastewater enters the system from the right compartment, the mixed liquor passes through the middle compartment and then enters the left compartment, which serves as the sedimentation tank. The water flow direction is reversed, and the operation process is the same. This process has been running in our plant for two years. I think there are some loopholes in this process in terms of phosphorus and nitrogen removal, that is, the water discharged from the sedimentation tank in each main stage has not gone through a complete anaerobic-aerobic process, and the discharged water is mainly aerobic water.

On the other hand, I think the anaerobic-aerobic time allocation in the current process is unreasonable, with the aerobic time being too long. Therefore, I put forward some suggestions. Taking the first main stage as an example: Wastewater first enters the left compartment for anaerobic stirring. After anaerobic stirring for a period of time, the wastewater is transferred to the middle compartment, and the left compartment stops anaerobic stirring and switches to aerobic aeration. In this way, the left compartment is "locked", so that the nitrification reaction can be completed as much as possible. After that, the left compartment stops aeration and serves as a sedimentation tank. Then, the second main operation stage is entered, and the wastewater flow direction is from right to left, and the operation process is the same.

After the suggestion was put forward, we also practiced it for a period of time. During the practice, we encountered such a problem, that is, when one compartment is "locked" for aeration, and the water is improved in the middle compartment, the sludge in the middle compartment is always pushed to the other compartment that serves as the sedimentation tank, resulting in extremely low sludge concentration in the middle compartment and very high sludge concentration in the sedimentation tank compartment, causing "sludge bulking" and secondary release of phosphorus.

For the above-described situations, I would like to ask the following questions:

(1) Is my suggestion reasonable for the current process of our plant? (2) Can the suggestion solve the problem of a large amount of sludge being pushed in the middle compartment? (3) Is the anaerobic-aerobic time allocation in the current process of our plant reasonable?

Answer: The answers to the three questions are as follows:

Your suggestion is more reasonable than the current operating mode. However, some adjustments need to be made. That is, while locking the left pool, extend the water inflow time of the left pool and reduce the water inflow time of the middle pool accordingly. This is more reasonable, and the reason is explained below.

After increasing the water inflow time of the left pool, more sludge in the left pool is pushed to the middle pool, resulting in more sludge in the middle pool than before the adjustment. This can increase the sludge concentration at the end of the middle pool's water inflow time compared to the current operating mode.

As for the anaerobic and aerobic time, it needs to be determined by trial and error based on the denitrification and phosphorus removal effect. Regardless of how the water inflow time of the left and middle pools is adjusted, the total water inflow time of the two pools remains unchanged. When the water inflow time of the middle pool increases and the water inflow time of the left pool decreases, the flow rate pushed to the right pool is the same, but the absolute amount of sludge flowing through will decrease. Of course, the sludge concentration in each pool cannot be balanced; this is a characteristic of the alternating aeration tank.

It is incorrect to shorten the cycle time. For processes with an anaerobic section, if the cycle time is shortened, the aerobic and anaerobic times in each cycle will be insufficient because the pre-sedimentation time before the side pool effluent cannot be shortened. Even without considering phosphorus removal, to shorten the cycle, it is necessary to have good sludge settling performance to reduce the pre-sedimentation time and ensure the time for the biochemical stage. It should also be noted that the UNITANK process has certain limitations on denitrification and phosphorus removal; phosphorus removal will restrict the denitrification effect.

29. How to count when performing microbial microscopic examination? I use a 10× objective lens and a 16× eyepiece, with a total magnification of 160×. Three bell animals are seen in one field of view under a total magnification of 160×. How many bell animals are there in 1 square centimeter?

Answer: 100× magnification should be used, i.e., both the eyepiece and the objective lens are 10×, to observe protozoa and metazoa and count them. The abundance of filamentous bacteria can also be roughly seen at 100×, while the sludge structure and the density of free bacteria are better observed at 400×. The counting method is as follows: First, determine how many drops there are per milliliter of aeration tank mixed liquid (assuming 20 drops per milliliter). Take one drop of the mixed liquid onto a slide, carefully cover it with a cover slip, then observe all the sludge samples under 100× magnification, record the number of various protozoa and metazoa, and then observe other contents.

30. Paper wastewater (wheat straw pulping) is being treated using a Carrousel oxidation ditch, but the sludge settleability in the oxidation ditch is currently very poor, and the SV30 is very poor. What is the reason for this?

Answer: The reason may be that in order to meet the oxygen supply, the aerator has to run at high speed, which breaks up the sludge and makes the settling performance even worse. This type of wastewater is suitable for the airlift aeration method, using a plug flow type. The current method is to avoid the aerator running at high speed for a long time as much as possible, control the sludge concentration, and keep the recycle ratio as small as possible to avoid the upward flow velocity in the sedimentation tank being too fast.

31. I believe that the side-channel sludge discharge of the three-compartment oxidation ditch has its advantages, but it also has fatal disadvantages, similar to the SBR process, which will form a sludge discharge funnel, resulting in a high concentration of sludge discharge in the initial stage and a very low concentration in the later stage. This is detrimental to the subsequent sludge treatment process and makes the control system complex, requiring unreliable instruments or increased labor intensity to complete.

Answer: This is completely avoidable. Side-channel sludge discharge is not possible at any time. If sludge is discharged from the aeration side channel during the A phase, this situation will not occur. Sludge with good settling performance does not necessarily require side-channel sludge discharge; it should be determined based on the specific situation of each device. As for convenient operation and management, there should be a reliable control system. The current control system is considered simple and mature. Of course, if the automatic control system has problems, manual control is very inconvenient, which is also one of the weaknesses of the three-compartment oxidation ditch.

32. How is the alternating sludge discharge of the three-compartment oxidation ditch performed? Is it switched based on the measured sludge concentration in the aeration tank or based on the predicted inflow concentration?

Answer: Sludge can be discharged from the aeration side channel at the beginning of the A and D phases. At this time, the sludge concentration in the aeration ditch is also relatively high. During the sludge discharge process, some substances adsorbed by the sludge can be discharged along with the sludge, which can also reduce the treatment load in the subsequent response phase. In short, the sludge discharge method and time need to be comprehensively considered based on the running cycle time, sludge settling performance, etc., and cannot be immutable. The alternating sludge discharge mode needs to be controlled by a separate control system, and the existing control program of the three-compartment oxidation ditch cannot meet this requirement.

33. How is the operating mode of the three-compartment oxidation ditch programmed? How is the operating time of each phase determined?

Answer: Since the operating states of the first three operating phases and the last three operating phases in a running cycle are the same, only the first three phases need to be considered when setting.

For example: The total time of the three phases A, B, and C is 4 hours. The time of phase C should be determined first. This phase is mainly for sedimentation. If the mixed liquid in the side channel used for sedimentation can completely separate the mud and water within 1 hour after stopping aeration, then the time of phase C is set to 1 hour;

Phase A is the main period of biochemical reaction, and its running time should be much longer than phase B. After the operation of phase A, most of the biochemical reactions have been largely completed;

Phase B is the transition phase from phase A to phase C. At this time, the wastewater enters the middle ditch, and after biochemical treatment, it flows to another sedimentation ditch. The aeration side ditch continues aeration without wastewater inflow, allowing the substances that have not yet been degraded in the ditch to be further converted. Therefore, the time of phase B is shorter. Different operating modes should be adopted according to different situations. For example, when the sludge settling performance is poor, the time of phase C should be appropriately increased, and the time of phases A and B should be reduced accordingly. If necessary, a transition phase can be set between C and D.

34. Our unit uses a Carrousel oxidation ditch 2000 type process for a municipal wastewater treatment plant with a scale of 80,000 tons/day. During operation, the removal of NH 3 -N is not ideal. In February, the influent NH 3-N averaged 32.35 mg/L, and the effluent was 25.99 mg/L. Can the value be reduced by increasing the DO value of the aerobic zone? NH 3-N ?

Answer: The dissolved oxygen in the aerobic zone can be increased, and the inner reflux gate can be opened wider, which reduces the anoxic volume of the denitrification zone and can improve the nitrification effect to a certain extent. In addition, factors such as whether the alkalinity is sufficient should also be considered.

35. The hydraulic design of the Carrousel oxidation ditch is still an insufficiently explored topic in China. I think the main reason is that it involves various factors:

Factors such as the mechanical and hydraulic performance of mechanical equipment (especially surface aerators), including the shape of aeration impellers, rotational speed, immersion depth, and the energy input into the water during operation (this energy is distributed among oxygenation, propulsion, and mixing); and the specific layout and design of the oxidation ditch, such as channel length, width, and depth, the position and shape of the guide walls, and whether they are eccentrically set.

Combining all these factors (and possibly others not mentioned above), we can determine the specific water flow patterns and parameters in the carrousel oxidation ditch (such as streamlines, turbulence intensity, cross-sectional velocity distribution, and average velocity). Given the complexity of this issue, what suggestions do you have regarding the hydraulic design of the carrousel?

Answer: It's not that complicated. The flow velocity in the oxidation ditch doesn't have a definite relationship with the hydraulic retention time or the volume of the oxidation ditch. The flow velocity in the oxidation ditch is controlled to prevent sedimentation. It shouldn't be too high; too low a velocity will cause sludge to settle. This is achieved using underwater propellers or surface aerators. The equipment used to achieve the desired flow velocity depends on the depth and length of the basin, and different manufacturers have different equipment selections.

36. Could you provide information on the precautions and limitations of operating and managing a three-ditch oxidation ditch?

Answer: There are many things to note. First, the operating cycle time must be determined based on the actual situation, and then the time for each operating stage within the cycle must be determined.

The C-stage duration should be determined first, as this is the sludge-water separation time. The immersion depth of the rotating brush should also be adjusted to ensure good oxygenation capacity and mixing power. The immersion depth of all rotating brushes in the basin should be consistent. The immersion depth of the rotating brush should be adjusted in a stationary state using the effluent weir gate, i.e., by raising and lowering the effluent weir gate when water enters the oxidation ditch but aeration is not yet started. When the rotating brush is at the appropriate immersion depth, the opening of the effluent weir gate is the limit for the rotating brush operation.

The limit for the opening of the effluent weir gates in the two side ditches should be basically the same. Based on the characteristics of the wastewater and the actual situation of the device, the optimal operating mode for daily operation should be determined through trial operation and input into the programmable logic controller (PLC) for operational control.

The operating mode should be adjusted promptly when abnormal situations occur. For example, if sludge sedimentation performance is poor, resulting in difficulty in sludge-water separation in the sedimentation ditch and muddy effluent, the time of the C stage should be increased, and the time of other stages should be reduced accordingly. The opening and closing of the effluent weirs in the two side ditches are automatically controlled according to the set process requirements. During the switching of the two side ditches in half a cycle, at the preset time, the weir gate of the original effluent ditch should be closed only after the weir gates of the other pre-sedimentation ditch are fully opened to prevent sludge from floating in the original pre-sedimentation ditch during the initial effluent time.

If the automatic control system malfunctions, manual control can be used. During manual control, the opening and closing time and sequence of each device should strictly follow the operating mode and be the same as the automatic control program. In the calculation of sludge load and sludge age, the volume of the biological part can be calculated as the total volume of the oxidation ditch multiplied by the ratio of the total biological time to the total hydraulic retention time.

37. Our wastewater treatment plant has been in operation for nearly six years. Sludge bulking has occurred in the past two months and cannot be effectively controlled. The process is ICEAS, with a settling ratio of 60 to 90%, but the filamentous bacteria are generally normal. The aeration time is generally controlled according to the dissolved oxygen in the water, stopping aeration when it reaches 5.0 to 5.6. The main pollutants in our plant are ethanol, which often causes instantaneous shock loads. Please advise.

Answer: This type of water easily causes bulking because of high soluble organic matter. N and P are insufficient and need to be added.

38. Our plant has two carrousel oxidation ditches, designed for a daily treatment capacity of 80,000 tons. Currently, only one system is operating, with a daily treatment capacity of 40,000 tons. The second system will be commissioned after the new year. We will use the sludge from the first system to cultivate the sludge in the second system. Please explain how to do this.

Answer: Since one system is already running, cultivation isn't necessary. You can just accumulate more sludge before starting up the other system and introduce it then.

39. From a practical perspective, discuss your views on the application of automatic control technology in the wastewater treatment industry, such as for the carrousel process.

Answer: There are many biochemical treatment processes. It depends on the specific process. If it's the traditional air-blown activated sludge process, automatic control isn't necessary. Manual remote control of the level protection control and pumps is sufficient.

Automatic control is certainly better for carrousel oxidation ditches. If there are underwater propellers, protective control is sufficient. If there are no underwater propellers, operational control is best. Here, protective control refers to the control system (such as a PLC) controlling the dissolved oxygen within the set range by starting and stopping the aerator and adjusting its speed. Operational control is different. In addition to the above requirements, it also considers preventing sludge sedimentation when the aerator is running slowly or only some aerators are running, i.e., when the overall operating state of the aerator only meets the DO control but not the sludge-water mixing, it can automatically adjust.

40. We are currently monitoring two basins. Basin 1 has sludge inoculation, but after one month of microscopic examination, only a large number of Paramecium were found, and the probability of finding Vorticella is basically zero, at most a few nematodes;

Basin 2 did not add inoculated sludge, and aeration was started after the inflow. After one month, microscopic examination revealed a large number of Vorticella and some Paramecium and other bacteria, but the sludge content in both basins is very low.

How can we cultivate the bacteria in Basin 1 and increase the sludge content? Also, the dissolved oxygen in the aeration tank is very high, generally between 9 and 11, rarely below 6, and only occasionally. Our blower is already running at the minimum, and the dissolved oxygen in Basin 2 is even higher, generally between 10 and 12.

Answer: The situation in both basins is similar, caused by insufficient nutrients and excessive aeration. The sludge is in a state of continuous growth and continuous self-oxidation. Therefore, the aeration time must be strictly controlled. If the sludge quantity cannot be increased, intermittent aeration can be adopted. Nutrient ratio control should also be considered.