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| Hangzhou AquaSust Water Technology Co., Ltd. Head ofice: #2808, Baiyue Center, Linping, Hangzhou, Zhejiang, China M:+86 152 6746 2807 Email: [email protected] website: www.nihaowater.com www.chinambbr.com www.aquasust.com www.aquasustfactory.com |
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| Sewage Treatment Air Flow Calculation | ||||
| Blue block is the design datameter: be filled in Brown:calculate process data Red:last result for your process |
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| 1.Sewage Treatment--Parameter | ||||
| Water volume m³/h | COD(mg/1) | BOD | ||
| 46 | 1200 | No data:BOD=0.5×COD=600mg/L | 600 | |
| 2. Calculation by Gas-Water Ratio | ||||
| Contact oxidation tank | 15:01 | 15×46=690m³/h | 690 | |
| Activated sludge tank | 10:01 | 10×46=460m³/h | 460 | |
| Conditioning tank | 5:01 | 5×46=230m³/h | 230 | |
| Total air volume | 690+460+230=1380m³/h=23m³/min | 1380 | ||
| 3. According to Remove 1kgBOD Need 1.5 Oxygen | ||||
| The amount of BOD removed per hour is: | 600mg/L×46m³/h=27.6kg | 27.6 | ||
| Oxygen demand | 27.6×1.5=41.4kg/h | 41.4 | ||
| Weight of oxygen in air | 0.233kgO2/kgAir | |||
| Density of air. | 1.293kg/m³ | |||
| Then the amount of air required: | 41.25÷0.233=177.7kg/h | 177.6824034 | ||
| Thenthe volume of air | 177.04÷1.293=137.4m³/h | 137.4187188 | ||
| Set the microporous aeration head oxygen utilization rate of 20% | ||||
| Thenthe actual amount of air required | 137.4÷0.2=687m³/h=11.45m³/min | 11.4515599 | ||
| 4. According to The Unit Pool Area Aeration Intensity Calculation | ||||
| Aeration intensity generally 10-20m³/m²h | ||||
| Aerobic pool to take the middle value of 15m³/m².h |
15 | 125 | ||
| Adjustment pool take 5m³/m².h | 3 | 120 | ||
| Contact oxidation and activated sludge tank area | 0.25 | |||
| Then the aerobic tank requires air volume | 125×15=1875m³/h=31.25m³/min | |||
| Thenthe air volume required for regulating tank | ||||
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