Water Treatment Questions and Answers (2)

6. What is the highest efficiency of contact oxidation treatment that you have seen or debugged? How much is the least? There should be a data range, right? Assuming that the effluent is a first-class standard, then the range of the concentration of organic matter in the influent comes out. Of course, this does not seem to be universal.

answer:

  1. During stable operation, the contact oxidation treatment efficiency is about 60% to 95%, which is related to the quality of raw water according to its position in the process.
  2. Usually in the treatment of industrial wastewater, especially tanning and dye wastewater, the treatment effect is low.
  3. The contact oxidation treatment is more suitable when dealing with the COD concentration of the influent water of about 1000-1500 ppm

7. What I am debugging now is the fluorescent whitening agent wastewater. The raw materials are mainly cyanuric chloride, DSD acid, aniline, diethanolamine, soda ash, p-aminobenzenesulfonic acid, etc. The current treatment process is a conditioning tank—micro-electrolysis– -UASB—aerobic a pool—sedimentation tank—-aerobic b pool—-second sedimentation tank. The water inflow is 3.5 cubic meters per hour. Due to the lack of aeration in the regulating tank, the UASB effluent fluctuates up and down, and the UASB effluent is unstable. The COD is between 1000 and 1800, and the chloride ion is about 9000 mg/L. After entering aerobic, tap water is added every hour. 2.5 cubic meters, add 75 kg of flour at the same time, two aerobic pools, each pool has an effective volume of 100 cubic meters, a small amount of bellworms can be seen in organisms, aerobic A pool sv42% (anaerobic effluent with mud, MLSS changes greatly) aerobic B The sv18% effluent of the pool is around COD650, the MLSS of the aerobic pool a is 815mg/L, the MLSS of the aerobic pool b is 216mg/L, the return sludge of the aerobic pool a is 675mg/L, and the return sludge of the aerobic pool b is 310mg/L. L, the aerobic A pool discharges 10 cubic meters of mud every day, and the mud age is estimated to be about 8 days. The aerobic B pool discharges 8 cubic meters of mud every day, and the mud age is about 9 days. The effluent is turbid. I feel that the sludge cannot be cultivated. The removal rate is not high, what is going on? How can I do this?

answer:

  1. It turned out to be a fluorescent whitening agent, which should belong to refractory organic substances. Therefore, it is not recommended to dilute the concentration by increasing tap water. In order to prolong the contact time in the aerobic tank, at the same time, the residence time of the wastewater in the UASB system should also be increased, so that the degradation can be completely degraded.
  2.  If conditions permit, strengthen the coagulation removal effect of the physicochemical section before entering the biochemical system.

8. The main process I debug at present: anoxic + aerobic + anoxic + aerobic, when the cod is high, the ammonia nitrogen removal effect is very poor. The influent ammonia nitrogen is 200 and cod is 1000. When the ammonia nitrogen inflow is 200 and the cod inflow is 200 some time ago, the ammonia nitrogen effluent effect is very good (40-50). The design value is 60, and the cod mainly uses waste methanol. At present, it has been debugged for one and a half months, and no sludge has grown. The measured sludge concentration is 2000-3000 and sv10-15. The company requires it to be adjusted as soon as possible, but what is the most appropriate way to do it now?

answer:

  1. MLSS already has 2000-3000ppm, please confirm the F/M value, and then see if it is necessary to increase its concentration.
  2. Your system is used to remove ammonia nitrogen in a targeted manner, and the concentration of organic matter is sufficient. Please confirm whether the hypoxia control of the hypoxia tank is in place, and whether the residence time of the water is too short.

9. In the design of wastewater treatment process, if hydrolytic acidification and UASB are designed, should hydrolytic acidification be placed behind or in front of UASB? Can you give a specific introduction?

Answer: It must be placed in the front. Placing the hydrolysis tank in the front can improve the biochemical properties of wastewater, remove some inorganic COD, and turn the refractory organic substances of macromolecules into small molecules, which are easy to degrade, and have a certain function of buffering high-load wastewater. Putting it behind the UASB pool is not really necessary.
In addition, the role of hydrolysis and acidification belongs to the category of pretreatment, and UASB can be used as a final treatment facility. In terms of its tolerance to influent water, the hydrolysis and acidification tank should be in the front. For example, when high SS wastewater flows in, the hydrolysis and acidification tank does not fluctuate much, but the UASB tank is unbearable, and the effluent fluctuates significantly.

10. The food-to-microbe ratio is the ratio of nutrients to microorganisms, but how to calculate it? Can you be more specific?

Answer: The calculation method of food micro ratio:
Food to microratio (F/M) is expressed by BOD-sludge load rate (Ns) in practical application.
Namely: Ns=(QLa)/(XV)(kgBOD5/kgMLSS•d)
where:
Q—Sewage flow (m3/d)
V—aeration volume (m3)
X—Mixed liquid suspended solids (MLSS) concentration (mg/l)
La—influent organic matter (BOD) concentration (mg/l)

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Further reading:

water treatment for filling water maker manufacturer bottling machine

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