Technical consultation: Questions regarding aeration volume
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The last edit to this post was made by hesonchang214 on 2009-12-21 at 21:29. Is the aeration volume in aerated tanks calculated using formulas by everyone? Also, how should the air volume be determined for aeration tanks such as activated sludge aeration tanks and the O tank in A2/O systems? And how should the capacity of the blowers be selected? .
Calculation of aeration volume
There are various methods for calculating the aeration volume. I tried to calculate it using different methods and found that there were significant differences. I’m sharing these calculations here for everyone’s feedback on which method is more accurate.
Parameters:
Water flow rate: 46 tons/hour; COD: 1200 mg/l; no data available for BOD. It is assumed that BOD = 0.5 × COD = 600 mg/l.
**Method 1: Calculation based on air-to-water ratio**
- For contact oxidation tanks: ratio of 15:1, so air volume = 15 × 46 = 690 m³/h.
- For activated sludge tanks: ratio of 10:1, so air volume = 10 × 46 = 460 m³/h.
- For the equalization tank: ratio of 5:1, so air volume = 5 × 46 = 230 m³/h.
Total air volume = 690 + 460 + 230 = 1380 m³/h = 23 m³/min.
**Method 2: Calculation based on the requirement of 1.5 kg of O2 per 1 kg of BOD removed**
- BOD removed per hour = 0.6 kg/m³ × 1100 m³/day ÷ 24 = 27.5 kgBOD/h.
- Oxygen required = 27.5 × 1.5 = 41.25 kgO2.
- The weight of oxygen in air is 0.233 kg O2/kg air. Therefore, air volume required = 41.25 kgO2 ÷ 0.233 kg O2/kg air = 177.04 kg air.
- With an air density of 1.293 kg/m³, the volume of air needed is 177.04 kg ÷ 1.293 kg/m³ = 136.92 m³.
- Given that the oxygen utilization efficiency of micro-porous aeration heads is 20%, the actual air volume required is 136.92 m³ ÷ 0.2 = 684.6 m³ = 11.41 m³/min.
**Method 3: Calculation based on aeration intensity per unit area of the tank**
- The typical aeration intensity is 10–20 m³/m²/h; taking the average value, the aeration intensity is 15 m³/m²/h.
- Total area of contact oxidation tanks and activated sludge tanks = 125.4 m². Thus, air volume = 125.4 × 15 = 1881 m³/h = 31.35 m³/min.
- Aeration intensity for the equalization tank is 3 m³/m²/h, with an area of 120 m²; thus, air volume = 3 × 120 = 360 m³/h = 6 m³/min.
- Total air volume required = 37.35 m³/min.
**Method 4: Calculation based on the number of aeration heads**
- The tank capacity is determined based on the retention time. It is calculated that 350 aeration heads are needed, with each head requiring 3 m³/h of air. Thus, total air volume required = 350 × 3 = 1050 m³/h = 17.5 m³/min.
- Adding the air volume required for the equalization tank, which is 6 m³/min, the total air volume required is 23.5 m³/min
There are various methods for calculating the aeration volume. I tried to calculate it using different methods and found that there were significant differences. I’m sharing these calculations here for everyone’s feedback on which method is more accurate.
Parameters:
Water flow rate: 46 tons/hour; COD: 1200 mg/l; no data available for BOD. It is assumed that BOD = 0.5 × COD = 600 mg/l.
**Method 1: Calculation based on air-to-water ratio**
- For contact oxidation tanks: ratio of 15:1, so air volume = 15 × 46 = 690 m³/h.
- For activated sludge tanks: ratio of 10:1, so air volume = 10 × 46 = 460 m³/h.
- For the equalization tank: ratio of 5:1, so air volume = 5 × 46 = 230 m³/h.
Total air volume = 690 + 460 + 230 = 1380 m³/h = 23 m³/min.
**Method 2: Calculation based on the requirement of 1.5 kg of O2 per 1 kg of BOD removed**
- BOD removed per hour = 0.6 kg/m³ × 1100 m³/day ÷ 24 = 27.5 kgBOD/h.
- Oxygen required = 27.5 × 1.5 = 41.25 kgO2.
- The weight of oxygen in air is 0.233 kg O2/kg air. Therefore, air volume required = 41.25 kgO2 ÷ 0.233 kg O2/kg air = 177.04 kg air.
- With an air density of 1.293 kg/m³, the volume of air needed is 177.04 kg ÷ 1.293 kg/m³ = 136.92 m³.
- Given that the oxygen utilization efficiency of micro-porous aeration heads is 20%, the actual air volume required is 136.92 m³ ÷ 0.2 = 684.6 m³ = 11.41 m³/min.
**Method 3: Calculation based on aeration intensity per unit area of the tank**
- The typical aeration intensity is 10–20 m³/m²/h; taking the average value, the aeration intensity is 15 m³/m²/h.
- Total area of contact oxidation tanks and activated sludge tanks = 125.4 m². Thus, air volume = 125.4 × 15 = 1881 m³/h = 31.35 m³/min.
- Aeration intensity for the equalization tank is 3 m³/m²/h, with an area of 120 m²; thus, air volume = 3 × 120 = 360 m³/h = 6 m³/min.
- Total air volume required = 37.35 m³/min.
**Method 4: Calculation based on the number of aeration heads**
- The tank capacity is determined based on the retention time. It is calculated that 350 aeration heads are needed, with each head requiring 3 m³/h of air. Thus, total air volume required = 350 × 3 = 1050 m³/h = 17.5 m³/min.
- Adding the air volume required for the equalization tank, which is 6 m³/min, the total air volume required is 23.5 m³/min
Another issue is the selection of air volume – how to determine the air volume of the blower? The issues related to aeration volume and fan selection have been plaguing me for almost two years; the problem is that I have no experience in biochemical processes, as the projects my company handles are all of a physical-chemical nature. I hope someone knowledgeable can answer! !
The authentic formula for calculation: AOR=a’QLr+b’VN. In the formula, AOR represents the designed oxygen demand (kgO2/d). a’ is the amount of oxygen required per kg of BOD for oxidation, typically ranging from 0.42 to 0.53. b’ is the oxygen demand for the oxidation of sludge itself, expressed as 1/a; it is in units of kgO2/kgMLVSS, with values generally ranging from 0.188 to 0.11. L denotes the concentration of BOD removed (kg/m), while Q represents the designed flow rate of the influent water (m/d)
Selection of blowers: It is primarily determined based on flow rate and pressure, while also taking into account other requirements such as installation and noise levels. Generally, the type is chosen first, such as centrifugal or axial flow. When selecting a model, the main considerations are flow rate, head pressure, whether blade cutting is required, and the size of the motor. Finally, meet other requirements such as noise.
This post was last edited by cdpulin on 2009-12-23 at 15:49. Industrial wastewater: calculated based on the water quality and the steam-to-water ratio. Domestic wastewater: Calculated using a formula. Of course, it can also be calculated according to the German ATI standards (applicable to both industrial and domestic use), but it may not be suitable for China’s national conditions.