Biological Wastewater Treatment Process - Interpretation of Aeration Capacity and Equipment
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- Mar 21,2025
Summary
Aeration and oxygen supply should be relatively simple for experts, and mainly involves three items: calculation of aeration volume (which has a lot to do with water quality and water volume), aeration fans and aeration devices. There are many methods to calculate aeration volume, each method has its own advantages and disadvantages (mainly simplicity and deviation rate). Everyone can choose a method according to the project situation and their own needs.
By: Kate Nana
Post Date:March 21, 2025
Post Tags:Biological Wastewater Treatment Process - Interpretation of Aeration Capacity and Equipment
Biological Wastewater Treatment Process - Interpretation of Aeration Capacity and Equipment
1. The role of aeration
Aeration has two main functions. One is mixing and stirring to accelerate the homogenization of wastewater (the equalization tank only has the function of homogenization, while the aeration tank has the function of preventing sludge sedimentation in addition to homogenization); the other is to provide food for aerobic microorganisms. Oxygen will participate in the metabolic activities of aerobic microorganisms. The following are the specific functions:
Oxygen supply
Mixing and stirring
Through the airflow generated by aeration or mechanical stirring, the sewage and activated sludge are fully mixed to ensure uniform contact between microorganisms and pollutants and improve treatment efficiency.
Prevent sludge sedimentation
Removal of volatile substances
2. Calculation method of aeration
The calculation methods are mainly divided into empirical formulas and theoretical formulas. The empirical formulas are empirical parameters obtained through multiple actual cases, which can be used as data verification or non-professional personnel calculation methods. Their main characteristics are simple to learn and fast calculation, but the numerical deviation is large. The theoretical formulas are two commonly used formulas in water treatment textbooks. Not to mention laymen, even experts will have a headache when seeing these two formulas. The formulas are relatively complex and involve more parameters (such as water temperature, oxygen saturation, altitude, water quality correction, etc.). The advantage is that they are relatively accurate. Of course, accuracy can also be a disadvantage. The actual water quality indicators of a sewage treatment plant have a certain fluctuation range, and the initially calculated values are not necessarily accurate data. The main purpose of aeration volume calculation is to have an approximate oxygen supply range, and then adjust and correct it based on the DO index of real-time monitoring at the actual site.
Reference data: The sewage treatment plant treats 200m³/d of water, COD is calculated at 500mg/L, and BOD is calculated at 300mg/L, which are similar to domestic sewage.
2.1 Calculated by gas-water ratio
The air-water ratio, that is, the ratio of aeration volume to sewage volume, for a certain industry or category of sewage, the ratio will be within a relatively certain range. The following ratio is based on domestic sewage as a reference.
For biochemical aeration tanks, the choice of process (whether contact oxidation or activated sludge, etc.) will basically not affect the gas supply. The gas supply is only related to the COD concentration (it is recommended to calculate the gas supply based on COD instead of BOD, because the final COD must be removed biochemically)
For aeration tanks, the air supply:
sewage volume = 10-15:1, and the middle value of 12.5:1 can be taken. In addition, for biochemical pools that use the MBR process, the gas-water ratio needs to reach 30-40:1, otherwise the hollow fiber membrane will easily become blocked.
For the equalization tank, the air supply volume:
sewage volume = 2-5:1, and the middle value of 3:1 can be taken. The equalization tank does not actually need to be aerated.
The gas supply volume is: (12.5+3)*200=3100m³/d=2.58m³/min
Without considering the regulating tank, it is 12.5*200=2500m³/d=2.1m³/min
Of course, this aeration volume takes into account the oxygen utilization rate of the aeration device. The calculation is very simple and can be calculated by a layman. However, this method is not recommended for the industry, but it can be used to calculate the accuracy of aeration volume.
Some people may ask, if it is not domestic sewage but other types of industrial wastewater, what value this ratio can be. It can be determined by the COD ratio. If the COD is within 500, the domestic sewage value can be used as a reference. If it exceeds 500, the ratio of this type of wastewater to 500 can be calculated, and then the value can be multiplied by the gas-water ratio of domestic sewage.
For example, if the COD of an industrial wastewater is 3000 mg/L, then its gas-water ratio is 3000/500*10:1=60:1.
2.2 Calculation by COD
This method can be used as a simple algorithm with reasonable accuracy. It ignores the oxygen demand for nitrification of ammonia nitrogen and only considers the oxygen demand for degradation of organic pollutants (COD). Generally speaking, COD is nominally the oxygen demand (you can check the COD
Definition), a certain margin should be reserved, the ratio of gas supply to COD can be 1.05:1. If ammonia nitrogen nitrification needs to be considered, the ratio of ammonia nitrogen nitrification oxygen demand to ammonia nitrogen is 4.57:1.
The daily COD removal is 0.5kg/m³×200m³/d=100kgCOD/d
The oxygen required is 100×1.05=105kgO2/d
Under standard conditions, the weight of oxygen in each cubic meter of air is 0.28kgO2/m3
The required air volume is 105/0.28=375m3/d
Biochemical ponds generally use microporous aeration heads, and their oxygen utilization rate is about 15-20%, with a value of 17.5%.
The required fan capacity is: 375/17.5%=2143m³/d=1.8m³/min
Check the gas-water ratio: 2143/200=10.7:1, which basically corresponds to 2.1.
2.3 Calculation of aeration intensity per unit pool area
It is not recommended, the error is huge!
The reasons are analyzed as follows:
① The volume of the biochemical pool will be different if the process is different. The volume load of different processes is quite different. The area of the biochemical pool will be different if the pool depth is different.
②The main influencing factor of aeration is the COD in the wastewater. For example, when the COD concentration is 500 or 3000, the aeration intensity will not be within the same range. For simple mixing and stirring, please refer to the aeration intensity of the equalization tank.
According to reference materials found on the Internet, the aeration intensity is generally 10-20m³/m²h. Taking the middle value, the aeration intensity is 15m³/m²h. Since it is necessary to calculate the capacity of the biochemical pond, the oxygen supply will not be calculated here.
There is also a method on the Internet that calculates the air supply by the number of aeration plates, which is also not recommended. This method itself is putting the cart before the horse. The normal order should be to calculate the air supply first, then calculate the number of aeration plates based on the air supply of a single aeration plate, and finally the plane layout of the aeration plates!
2.4 Calculation based on theoretical formula (Section 7.9 of the outdoor drainage design standard)
This formula is broken down into 4 steps:
When carbon-containing substances are measured as BOD5, the oxygen equivalent of carbon is 1.47. (It is more accurate to use COD as the oxygen equivalent of carbon 1.05)
Reference data: sewage plant inlet flow rate 200m³/d, inlet COD 500mg/L, BOD 300mg/L, effluent BOD 30mg/L, residual sludge concentration MLVSS=2500g/m3, inlet Kjeldahl nitrogen 50mg/L, effluent Kjeldahl nitrogen 20mg/L, inlet total nitrogen 55mg/L, effluent nitrate nitrogen 10mg/L.
Overall formula:
Oxygen demand of sewage = oxygen demand for carbon oxidation - oxygen demand for microbial metabolism + oxygen demand for ammonia nitrogen nitrification - oxygen demand for denitrification
Disassembly 1:
Carbon oxidation oxygen demand = 0.001*1.47*treated water volume*(influent BOD5-effluent BOD5)
Oxygen demand for carbon oxidation = 0.001*1.47*200*(300-30) = 50kgO2/d
Referring to Section 2.2, it is found that the data has a large deviation. Calculated by COD, assuming that the effluent COD is 50.
Oxygen demand for carbon oxidation = 0.001*1.05*200*(500-50) = 94.5kgO2/d
Disassembly 2:
Oxygen demand for microbial metabolism = 1.42 * daily emission of microorganisms
The amount of microorganisms discharged daily = sludge coefficient 0.5 * wastewater BOD concentration 0.3 * wastewater volume 200m3/d = 3kg/d
Then the oxygen demand of microorganisms' own metabolism = 1.42*3 = 4.3 kgO2/d
Therefore, some values are relatively small, and this part can be ignored in the actual calculation of aeration volume!
Disassembly 3:
Ammonia nitrogen nitrification oxygen demand = oxygen required to oxidize each kilogram of ammonia nitrogen * [0.001 * treated water volume * (influent Kjeldahl nitrogen - effluent Kjeldahl nitrogen) - MLVSS nitrogen content]
If there is usually no Kjeldahl nitrogen data, ammonia nitrogen data can be used instead!
Ammonia nitrogen nitrification oxygen demand = 4.57*[0.001*200*(50-20)-0.12*3]≈25.8kgO2/d
Disassembly 4:
Oxygen demand for denitrification = 0.62* oxygen required to oxidize one kilogram of ammonia nitrogen*[0.001* treated water volume*(influent total nitrogen-effluent Kjeldahl nitrogen-effluent nitrate nitrogen)-MLVSS nitrogen content]
If there is no Kjeldahl nitrogen data, total nitrogen data can be used instead of nitrate nitrogen!
Oxygen demand for denitrification = 0.62*4.57[0.001*200*(55-20-10)-0.12*3]≈13.2kgO2/d
Overall formula:
Oxygen demand of sewage = oxygen demand for carbon oxidation - oxygen demand for microbial metabolism + oxygen demand for ammonia nitrogen nitrification - oxygen demand for denitrification = 94.5-4.3+25.8-13.2=102.8kgO2/d
In actual engineering, this formula can also be used for simple calculations by referring to Section 2.2, that is, only the oxygen demand for carbon oxidation and the oxygen demand for ammonia nitrogen nitrification are calculated.
Under standard conditions, the weight of oxygen in each cubic meter of air is 0.28kgO2/m3
The required air volume is 102.8/0.28=367m3/d
Biochemical ponds generally use microporous aeration heads, and their oxygen utilization rate is about 15-20%, with a value of 17.5%.
The required fan capacity is: 367/17.5%=2097m³/d=1.75m³/min
Check the gas-water ratio: 2097/200=10.5:1, which basically corresponds to Section 2.1.
From this we can see that the oxygen demand for microbial metabolism itself can be ignored. When the ammonia nitrogen concentration is not high, the oxygen demand for ammonia nitrogen nitrification and denitrification can also be ignored.