1, Calculation of aeration volume of membrane tank

The effect of MBR membrane aeration is flushing the surface of the membrane. Here， the amount of aeration is just for the aeration of the membrane module tank rather than the aerobic tank. The aeration hole is disposed directly under the membrane frame,flowing the mixed liquid up, reducing membrane fouling.

The amount of aeration required, for membrane washing, is much greater than the amount of biochemical effects. The required amount of gas is calculated: the amount of aeration of the membrane tank.

For example, in a 60 membrane module elements, the internal cross-sectional area of the membrane box is 3.53 m2, the aeration volume of a single membrane module is 8.8 m3/min, and the total aeration volume of the 12 membrane modules is 105.6 m3/min, and the wind pressure is according to the water depth. It is determined that the blower pressure of 5m water depth is selected to be 60KPa.

2) Calculation of aeration volume of aerobic tank

The aerobic tank is a structure for sewage treatment using an activated sludge method. The tank provides a certain SRT to meet the DO of the aerobic microorganisms and provide the mixing conditions of the sewage and the activated sludge.

Aeration is a means of bringing air into contact with water. The purpose is to dissolve oxygen into the water.

There are many methods for calculating the amount of aeration.

The following three methods are mainly used.

Parameters: Assume that the designed water volume is 416m3/h, COD is 1200mg/L, BOD5=0.5xCOD=600mg/L

1: Calculate: according to the ratio of gas to water

The rate of gas-water is 15:1, the air volume is 15×416=6240m3/h, and the total air volume is 6240/60=104m3/min.

2: Calculate: according to the removal of 1Kg BOD5 need 1.5kg O2

The amount of BOD5 removal per hour is 0.6KG/M3x416m3/h=249.6Kg BOD/h, requiring oxygen 249.6×1.5=374.4kg O2.

The oxygen content in the air is 0.238kg O2/kg, and the air requirement per hour is 374.4/0.233 =1606.87 Kg.

The density of air is 1.293kg/m3, then the air volume is 1606.87kg÷1.293kg/m3=1242.74m3

The oxygen utilization rate of the microporous aeration hole is 20%, and the actual required air volume per hour is 1242.74m3/0.2=6213.72m3= 103.56m3/min

3: Calculate: according to the number of aeration hole

Calculate the pool capacity according to the SRT, calculate the area of the pool. Generally an aeration head of 0.5m2. Calculating the total number of aeration heads to be 3150 according to the area of the pool, and the required gas volume is 2m2/(h* piece), then air is required : 3150*2=6300m3/h=105m3/min.

Generally, the amount of aeration is determined according to the number of aeration heads.

（from: The new process design MBR）