MBR membrane bioreactor and design parameters
Based on the principle of organic matter pollution, significant alkaline cleaning effect, and significant salt scaling through acid cleaning effect, the chemically enhanced anti cleaning program is introduced into the MBR membrane process. By using a low intensity chemical cleaning operation, the contamination of the MBR membrane is eliminated in the early stage of formation, preventing the membrane from not being able to recover and deteriorate in a timely manner.
4.1 Calculate air volume based on biological requirements.
① Remove the nitrogen content of the object.
The nitrogen concentration in residual sludge is 6%, and the amount of residual sludge generated is 45% of the inflow into BOD (actual experimental results). Therefore, remove NR from the object that contains nitrogen concentration.
NR=50 (total nitrogen of water supply) -200 x 0.45 x 0.06 (nitrogen content of residual sludge)=44.6mg/L.
The nitrogen concentration NR of the removal object is set to full nitrification, and the circulating liquid (return) amount is 400%. The nitrogen removal amount of the denitrification tank is
NR x (4/5) (Total nitrogen removal rate 80%)
② Oxygen required for BOD oxidation.
Remove the remaining amount of BOD after removing BOD from the denitrification tank. The unit BOD amount accompanied by NO3 denitrification consumption is α ( α= 2.8kg BOD/kg-N (for every 1kg of nitrogen removed, 2.8kg BOD is required), with a daily average sewage volume of q (m3/d), accompanied by the amount of BOD eliminated through denitrification.
Q × 200mg/L (inlet BOD value) – Q × α × NR × (4/5) (BOD value consumed by anaerobic denitrification)
If the unit oxygen required for BOD oxidation is set to a (a=0.5 kg O2/kg BOD) (0.5 kg BOD is required for every 1 kg BOD removed), then the oxygen required for BOD oxidation.
A x (Q x 200mg/L-Q x α * NR * (4/5) ÷ 1 000… 1)
③ The amount of oxygen required for nitrogen oxidation.
NH4 2O2 → NO3- H2O 2H.
The unit oxygen required for nitrogen nitrification is set to b (b=64/14), and the oxygen required for nitrogen oxidation is.
Q x NR x b ÷ 1000… 2.
④ The amount of oxygen required for internal respiration of sewage sludge.
The capacity of the aeration tank is V (m3), the unit oxygen consumption for internal respiration of the sludge is c (c=0.07g-O2/g-VSS), and the MLVS/MLSS ratio is 0.7 based on actual test results. Therefore, the required oxygen for internal respiration of the sludge is.
V x 20kg/m3 (sludge concentration MLSS) x c x 0.7… 3)
⑤ Oxygen required for aeration tank.
According to 1), 2), and 3).
Q x 0.5 x (200-100) ÷ 1000; Q x 0.206V x 0.98=0.256; Q x 0.98V.
⑥ Air volume required.
1m3 of air contains 0.277kg of oxygen, with an oxygen dissolution efficiency of 3% (important parameter)
(0.256Q0.98V) ÷ 0.277 ÷ 0.03=30.81Q18.0V.
4.2 Calculate the air volume using membrane components.
To ensure the required air flow rate on the membrane surface during the operation of the membrane separation device, each membrane support is set at 11-12L/min (flat plate membrane)
4.3 Final air volume.
Take the maximum values of 4.1 and 4.2.